`bb1)`

$\sqrt{x^2 - 4x + 6} = x + 4$ $($ĐKXĐ: $x \ge -4)$

$x^2 - 4x + 6 = (x + 4)^2$

$x^2 - 4x + 6 = x^2 + 8x + 16$

$-4x - 8x = 16 - 6$

$-12x = 10$

$x = -\dfrac{10}{12}$

$x = -\dfrac{5}{6} \text{ (TMĐK)}$

Vậy $x = -\dfrac{5}{6}$

`bb2)`

$\sqrt{x^2 - 3x + 2} - 3 - x = 0$

$\sqrt{x^2 - 3x + 2} = x + 3$

ĐKXĐ: $\left[\begin{matrix} -3 \le x \le 1 \\ x \ge 2 \end{matrix}\right.$

$x^2 - 3x + 2 = (x + 3)^2$

$x^2 - 3x + 2 = x^2 + 6x + 9$

$-3x - 6x = 9 - 2$

$-9x = 7$

$x = -\dfrac{7}{9} \text{ (TMĐK)}$

Vậy $x = -\dfrac{7}{9}$

`bb3)`

ĐKXĐ: $\left[\begin{matrix} x \ge 2 \\ x \le -2 \end{matrix}\right.$

$(x - 3)\sqrt{x^2 - 4} = x^2 - 9$

$(x - 3)\sqrt{x^2 - 4} = (x - 3)(x + 3)$

$(x - 3)\sqrt{x^2 - 4} - (x - 3)(x + 3) = 0$

$\Rightarrow (x - 3)\left(\sqrt{x^2 - 4} - (x + 3)\right) = 0$

$\text{TH1:}$

$x - 3 = 0$

$x = 3 \text{ (TMĐK)}$

$\text{TH2:}$

$\sqrt{x^2 - 4} - (x + 3) = 0$

$\Rightarrow \sqrt{x^2 - 4} = x + 3$ $(x \ge -3)$

Kết hợp ĐKXĐ ban đầu ta có: $-3 \le x \le -2$ hoặc $x \ge 2$

Bình phương `2` vế pt:

$x^2 - 4 = (x + 3)^2$

$x^2 - 4 = x^2 + 6x + 9$

$-4 - 9 = 6x$

$6x = -13$

$x = -\dfrac{13}{6} \text{ (TMĐK)}$

Vậy $x \in \left\{-\dfrac{13}{6}; 3\right\}$

`bb4)`

$x^2 - 7x + \sqrt{x^2 - 7x + 8} = 12$ $($ĐKXĐ: $x^2 - 7x + 8 \ge 0)$

$(x^2 - 7x + 8) + \sqrt{x^2 - 7x + 8} - 8 - 12 = 0$

$(x^2 - 7x + 8) + \sqrt{x^2 - 7x + 8} - 20 = 0 (*)$

Đặt $t = \sqrt{x^2 - 7x + 8}$ ($t \ge 0$)

Từ đó phương trình $(*)$ trở thành:

$t^2 + t - 20 = 0$

$t^2 + 5t - 4t - 20 = 0$

$t(t + 5) - 4(t + 5) = 0$

$(t - 4)(t + 5) = 0$

$\text{TH1:}$

$t - 4 = 0$

$t = 4 \text{ (TMĐK)}$

$\Rightarrow \sqrt{x^2 - 7x + 8} = 4$

$x^2 - 7x + 8 = 16$

$x^2 - 7x + 8 - 16 = 0$

$x^2 - 7x - 8 = 0$

$(x - 8)(x + 1) = 0$

$\text{TH1. 1:}$

$x - 8 = 0$

$x = 8 \text{ (TMĐK)}$

$\text{TH1. 2:}$

$x + 1 = 0$

$x = -1 \text{(TMĐK)}$

$\text{TH2:}$

$t + 5 = 0$

$t = -5 \text{(L)}$

Vậy $x \in \{-1; 8\}$

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